in a) the force pulling down is gravity, and since it is at rest, tension must be the equal and opposite force
F = ma a in this case is gravity ... 9.8
so, F = mg = Tension
But, the deriv is not unique to 1 equation,
and isn't the C somewhat arbitrary
for example:
Seperation of variables:
dy/y^(1/2) = 4x dx
Integrate :
2 y^(1/2) = 2x^2 +C1
Simplify
y^(1/2) = x^2 + C2 <--- and the deriv of this is still dy/dx = 4x*y^(1/2), then all I did...
One of the problems on my AP Calc test:
The point (1,9) lies on the graph of an equation y=f(x) for which dy/dx = 4x*y^(1/2) where x> or = to 0 and y > or = 0
When x=0 y=?
Seperation of variables:
dy/y^(1/2) = 4x dx
Integrate :
2 y^(1/2) = 2x^2 +C now, if you do C now:
2...
In class today we were doing a problem where we were given the graph of f (the derivitive of F).
The domain was -3<x < or = 8
f(8)=0 and f had changed from positive <8 to 0, indicating a probable relative max (the function looked like a sinusoid, but the points after 8 were not graphed...